# Quotient cancellation for invertible ideals of orders in quadratic fields

Let $$K/\mathbb{Q}$$ be an imaginary quadratic field, $$m\ge 1$$ be a positive integer and let $$\mathcal{O}=\mathbb{Z}+m\mathcal{O}_K\subset \mathcal{O}_K$$ be the unique order of index (equivalently, conductor) $$m$$. I want to know whether the "quotient cancellation property", by which I mean the existence of an isomorphism (as $$\mathcal{O}$$-modules)

$$\frac{\mathfrak{a}\mathfrak{b}}{\mathfrak{a}\mathfrak{c}}\simeq \frac{\mathfrak{b}}{\mathfrak{c}},$$ holds whenever $$\mathfrak{a}$$, $$\mathfrak{b}\supset \mathfrak{c}$$ are invertible fractional $$\mathcal{O}$$-ideals.

I am able to show it when the quotient $$\frac{\mathfrak{b}}{\mathfrak{c}}$$ has order relatively prime to $$m$$. In this case, one may scale both $$\mathfrak{b}$$ and $$\mathfrak{c}$$ by some $$\mu\in K^{\times}$$ to get integral $$\mathcal{O}$$-ideals which are both relatively prime to the conductor $$m$$ of $$\mathcal{O}$$. Doing the same for $$\mathfrak{a}$$, we can thus assume $$\mathfrak{a}$$, $$\mathfrak{b}$$ and $$\mathfrak{c}$$ to be integral, prime-to-$$m$$, $$\mathcal{O}$$-ideals.

Notice that one has a multiplicative bijection between the monoid of integral, prime-to-$$m$$, $$\mathcal{O}_K$$-ideals and the monoid of integral, prime-to-$$m$$, $$\mathcal{O}$$-ideals (as shown in Cox's Primes of the form $$x^2+ny^2$$, Proposition 7.20 or K. Conrad's note The conductor ideal, Theorem 3.8) given by intersection with $$\mathcal{O}$$ on one side and by multiplication with $$\mathcal{O}_K$$ on the other side. Making use of this, and of the "quotient cancellation property" available in the maximal order $$\mathcal{O}_K$$ (see this topic Cancellation in quotient of fractional ideals for instance), one can get the job done.

However when $$|\frac{\mathfrak{b}}{\mathfrak{c}}|$$ is no longer prime to $$m$$, one cannot multiply $$\mathfrak{b}$$ and $$\mathfrak{c}$$ by the same scalar to get prime-to-$$m$$ integral $$\mathcal{O}$$-ideals and something else is needed to prove (or disprove) the quotient cancellation property.

A related question is the following: if $$N$$ is square-free with $$(N,m)\neq 1$$, can we find examples of such invertible fractional $$\mathcal{O}$$-ideals $$\mathfrak{b}\supset \mathfrak{c}$$ such that $$\mathfrak{b}/\mathfrak{c}\simeq \mathbb{Z}/N\mathbb{Z} \,\, ?$$

• Anyone, any idea? – Yoël Feb 19 '19 at 8:07
• If $\mathfrak{a}, \mathfrak{b}, \mathfrak{c}$ are invertible, isn't $\mathfrak{a} \mathfrak{b}/\mathfrak{a} \mathfrak{c} = \mathfrak{b} \mathfrak{c}^{-1} \mathfrak{a}\mathfrak{a}^{-1}$ equal to $\mathfrak{b} \mathfrak{c}^{-1}$? – Bart Michels Sep 17 '19 at 9:10